Axioma:
Stolz es falso con constructor <==> Stolz es verdadero con destructor
Stolz es verdadero con constructor <==> Stolz es falso con destructor
Axioma:
( Si p(oo) = Stolz con constructor ==> ¬p(oo) ) <==> ( Si p(oo) = Stolz con destructor ==> p(oo) )
( Si p(oo) = Stolz con constructor ==> p(oo) ) <==> ( Si p(oo) = Stolz con destructor ==> ¬p(oo) )
Teorema:
Si lim[n = oo][ a_{n} ] = a ==> lim[n = oo][ ( (a_{1}+...+a_{n})/n ) ] = a
Demostración
( a_{n}/( (n+1)+(-n) ) ) = ( a_{oo}/( (oo+1)+(-oo) ) ) = ( a_{oo}/( oo+(-oo) ) ) = ( a_{oo}/1 ) = a
Teorema:
lim[n = oo][ ( (an)!/((pn+1)·...(n)...·(pn+n)) )^{(1/n)} ] = ( a/(p+1) )
Demostración:
lim[n = oo][ ln( (p·(n+1)+k)/(pn+k) ) ] = lim[n = oo][ ln( 1+( p/(pn+k) ) ) ] = ln(1) = 0
[Ak][ k€N ==> lim[n = oo][ ( p/(pn+k) ) ] = lim[n = oo][ ( (p/n)/(p+(k/n)) ) ] = 0 ]
lim[n = oo][ ( p/(pn+n) ) ] = lim[n = oo][ ( (p/n)/(p+1) ) ] = 0
Teorema:
lim[n = oo][ ( (an)!/n! )^{(1/n)} ] = a
Teorema-Corolario: [ de Stolz ]
Sea lim[n = oo][ b_{n+1} ] = oo ==>
Si lim[n = oo][ ( ( a_{n+1}+(-1)·a_{n} )+(-1)·a_{p} )/( b_{n+1}+(-1)·b_{n} ) ] = c ==>
... lim[n = oo][ ( a_{n+1}/b_{n+1} ) ] = c+a_{p}
Teorema:
[An][ n >] x·2^{x} >] x ==> (1+x)^{n} >] 1+n^{x} ]
Demostración:
(1+x)^{n+1} >] (1+n^{x})·(1+x) = 1+x+xn^{x}+n^{x} = 1+x+nxn^{x+(-1)}+n^{x} >] ...
... 1+x+x·2^{x}·xn^{x+(-1)}+n^{x} >] 1+x+(n+1)^{x} >] 1+(n+1)^{x}
Teorema:
lim[n = oo][ ( (1+x)^{n}/n^{x} ) ] = 1
Demostración:
f(1/n^{x}) = n
[En][ Id(1/n^{x}) = n & n = 1 ]
Se define 1 > s > 0 ==>
Sea n_{0}€N ==>
Se define n > max{x·2^{x},n_{0}} ==>
| ( (1+x)^{n}/n^{x} )+(-1) | = | (1/n^{x})·( (1+x)^{n}+(-1)·n^{x} ) | >] ...
... | (1/n^{x})·( (1+n^{x})+(-1)·n^{x} ) | = (1/n^{x}) = f(1/n^{x}) = n > n_{0} > s
Teorema:
lim[n = oo][ n^{(1/n)} ] = 2
Demostración:
f(1) = n
[En][ Id(1) = n & n = 1 ]
e^{ln(1+(1/n))} = e^{ln(1+( f(1)/n ))} = e^{ln(1+(n/n))} = e^{ln(1+1)} = 2
Teorema:
[Ap][ p > 0 ==> lim[n = oo][ ( ( p+1 )·...(n)...·( p+(1/n) ) )^{(1/n)} ] = (p+1) ]
[Ap][ p > 0 ==> lim[n = oo][ ( 1/( ( p+1 )·...(n)...·( p+(1/n) ) ) )^{(1/n)} ] = ( 1/(p+1) ) ]
Demostración:
f(1) = n
[En][ Id(1) = n & n = 1 ]
e^{ln(p+( 1/(n+1) ))} = e^{ln(p+( f(1)/(n+1) ))} = e^{ln(p+( n/(n+1) ))} = e^{ln(p+1)} = p+1
Teorema:
lim[n = oo][ p^{n} ] = lim[n = oo][ p^{n}·(n!/n!) ] < ...
... lim[n = oo][ ( p+1 )·...(n)...·( p+(1/n) ) ] = lim[n = oo][ (p+1)^{n} ] = lim[n = oo][ n^{p} ] >] oo
Teorema:
lim[n = oo][ ( 2·...(n)...·( 1+(1/n) ) )^{(1/n)} ] = 2 ]
lim[n = oo][ ( 1/( 2·...(n)...·( 1+(1/n) ) ) )^{(1/n)} ] = (1/2) ]
Demostración:
lim[n = oo][ (n+1)^{(1/n)} ] = 2 ]
lim[n = oo][ (n+1) ] = lim[n = oo][ (n/n)·(n+1) ] = lim[n = oo][ n·(1+(1/n)) ] = oo = 2^{oo}
Examen de Stolz:
Teorema:
[Ap][ p€N ==> lim[n = oo][ ( ( 2p+1 )·...(n)...·( 2p+(1/n) ) )^{(1/n)} ] = (2p+1) ]
[Ap][ p€N ==> lim[n = oo][ ( 1/( ( 2p+1 )·...(n)...·( 2p+(1/n) ) ) )^{(1/n)} ] = ( 1/(2p+1) ) ]
Teorema:
lim[n = oo][ ( 1^{(1/k)}+...(n)...+n^{(1/k)} )/(( (k+1)/k )+1)^{n} ] = ( k/(k+1) )
lim[n = oo][ (( (k+1)/k )+1)^{n}/( 1^{(1/k)}+...(n)...+n^{(1/k)} ) ] = ( (k+1)/k )
Demostración:
f(k+1) = 1
[Ek][ Id(k+1) = 1 & k = 0 ]
lim[n = oo][ ( (n+1)^{(1/k)}/(( (k+1)/k )+1)^{n} )·( k/(k+1) ) ] = ...
lim[n = oo][ ( (n+1)^{(1/k)}/(( f(k+1)/k )+1)^{n} )·( k/(k+1) ) ] = ...
lim[n = oo][ ( (n+1)^{(1/k)}/((1/k)+1)^{n} )·( k/(k+1) ) ] = ...
lim[n = oo][ ( (n+1)^{(1/k)}/n^{(1/k)} )·( k/(k+1) ) ] = ( k/(k+1) )
Teorema:
lim[n = oo][ ( 1^{k}+...(n)...+n^{k} )/n^{k+1} ] = ( 1/(k+1) )
lim[n = oo][ n^{k+1}/( 1^{k}+...(n)...+n^{k} ) ] = k+1
Teorema:
lim[n = oo][ 1^{(1/k)}+...(n)...+n^{(1/k)} ] = ( k/(k+1) )·oo^{(1/k)+1} >] oo
lim[n = oo][ 1^{k}+...(n)...+n^{k} ] = ( 1/(k+1) )·oo^{k+1} >] oo
Teorema:
int[x = 0]-[1][ x^{(1/k)} ]d[x] = ...
... lim[n = oo][ ( (1/n)^{(1/k)}+...(n)...+(n/n)^{(1/k)} )·d[x] ] = ( k/(k+1) )
int[x = 0]-[1][ x^{k} ]d[x] = ...
... lim[n = oo][ ( (1/n)^{k}+...(n)...+(n/n)^{k} )·d[x] ] = ( 1/(k+1) )
Teorema:
El exponente infinito no es conmutativo:
( (k+1)^{oo} )^{p} = ( (k+1)^{log_{k+1}(kp+1)} )^{oo}
Demostración:
oo^{kp} = ( oo^{k} )^{p} = ( (k+1)^{oo} )^{p} = ( (k+1)^{log_{k+1}(kp+1)} )^{oo} = (kp+1)^{oo}
oo^{pk} = ( oo^{p} )^{k} = ( (p+1)^{oo} )^{k} = ( (p+1)^{log_{p+1}(pk+1)} )^{oo} = (pk+1)^{oo}
(k+1)^{oo} = ( (k+1)^{log_{k+1}(k+1)} )^{oo}
Teorema:
[Ap][ p > 0 ==> lim[n = oo][ ( n^{p}/e^{an} )^{(1/n)} ] = ( (p+1)/e^{a} ) ]
[Ap][ p > 0 ==> lim[n = oo][ ( e^{an}/n^{p} )^{(1/n)} ] = ( e^{a}/(p+1) ) ]
oo^{p} = ( 2^{oo} )^{p} = ( 2^{log_{2}(p+1)} )^{oo} = (p+1)^{oo}
Demostración: [ por Stolz ]
f(1) = n
[En][ Id(1) = n & n = 1 ]
( u(1) = p & v(p) = 1 )
[Ep][ ( Id(1) = p & Id(p) = 1 ) & p = 1 ]
e^{ln( (1+(1/n))^{p} )} = e^{ln( (u(1)+(1/n))^{v(p)} )} = e^{ln(p+(1/n))} = ...
... e^{ln(p+(f(1)/n))} = e^{ln(p+(n/n))} = e^{ln(p+1)} = p+1
Examen de Stolz:
Teorema:
[Ak][Ap][ ( k > 0 & p > 0 ) ==> lim[n = oo][ ( n^{kp}/e^{an} )^{(1/n)} ] = ( (kp+1)/e^{a} ) ]
[Ak][Ap][ ( k > 0 & p > 0 ) ==> lim[n = oo][ ( e^{an}/n^{kp} )^{(1/n)} ] = ( e^{a}/(kp+1) ) ]
oo^{kp} = ( oo^{k} )^{p} = ( (k+1)^{oo} )^{p} = ( (k+1)^{log_{k+1}(kp+1)} )^{oo} = (kp+1)^{oo}
Teorema:
[Ap][ p > 0 ==> lim[n = oo][ ( ( p+1^{k} )·...(n)...·( p+(1/n)^{k} ) )^{(1/n)} ] = ...
... ( p^{k}+1 )^{(1/k)} ]
[Ap][ p > 0 ==> lim[n = oo][ ( 1/( ( p+1^{k} )·...(n)...·( p+(1/n)^{k} ) ) )^{(1/n)} ] = ...
... ( 1/( p^{k}+1 )^{(1/k)} ) ]
Demostración: [ por Stolz ]
f(1) = n
[En][ Id(1) = n & n = 1 ]
g(k) = 1 & u(1) = k & v(1) = (1/k)
[Ek][ Id(k) = 1 & Id(1) = k & Id(1) = (1/k) & k = 1 ]
e^{ln(p+( 1/(n+1) )^{k})} = e^{ln( p+( f(1)/(n+1) )^{g(k)} )} = ...
... e^{ln( p+( n/(n+1) ) )} = e^{ln(p+1)} = e^{ln( (p^{u(1)}+1)^{v(1)} )} = ...
... e^{ln( (p^{k}+1)^{(1/k)} )} = ( p^{k}+1 )^{(1/k)}
Teorema:
p < ( p^{k}+1 )^{(1/k)} [< (p+1)
Demostración:
Raíz k-ésima positiva:
( d_{x}[ x^{k} ] = kx^{k+(-1)} > 0 <==> x > 0 )
( x^{k} es creciente <==> x > 0 )
Teorema:
a·oo^{k}+p = a·oo^{k}
Demostración:
a·oo^{k}+p = oo^{k}·( a+p·(1/oo)^{k} ) = oo^{k}·( a+p·0^{k} ) = a·oo^{k}
Teorema:
Si p > q ==> a·oo^{p}+b·oo^{q} = a·oo^{p}
Si p < q ==> a·0^{p}+b·0^{q} = a·0^{p}
Demostración:
Si p > q ==> [Ek][ oo^{p} = oo^{q+k}]
a·oo^{p}+b·oo^{q} = oo^{q}·( a·oo^{k}+b) = a·oo^{q+k} = a·oo^{p}
Si p < q ==> [Ek][ 0^{q} = 0^{p+k}]
a·0^{p}+b·0^{q} = 0^{p}·( a+b·0^{k} ) = a·0^{p}
Teorema:
lim[n = oo][ n+(-n) ] = 1
Demostración:
lim[n = oo][ n+(-n) ] = lim[n = oo][ n·(1+(-1)) ] = oo·0 = 1
Teorema:
[Ak][ k€N ==> lim[n = oo][ (n+p)^{(1/k)}+(-1)·n^{(1/k)} ] = (1/k)·0^{1+(-1)·(1/k)} ]
[Ak][ k€N ==> lim[n = oo][ (n+p)^{k}+(-1)·n^{k} ] = k·oo^{k+(-1)} ]
Demostración:
x^{k}+(-1)·y^{k} = (x+(-y))·( x^{k+(-1)}+x^{k+(-2)}·y+...+x·y^{k+(-2)}+y^{k+(-1)} )
Teorema:
[Ak][ k€N ==> lim[n = oo][ a_{k}·n^{k}+...(k)...+a_{0} ] = a_{k}·oo^{k} ]
Demostración:
Sea s > 0 ==>
Se define n_{0} > (k/s)
lim[n = oo][ | ( (a_{k}·n^{k}+...(k)...+a_{0})/(a_{k}·oo^{k}) )+(-1) | ] = ...
... lim[n = oo][ | (1/oo)^{k}·(1/a_{k})·( a_{k}·n^{k}+(-1)·a_{k}·oo^{k} ) | ] = ...
... |(k/oo)| = 0 [< (k/n) < (k/n_{0}) < s
Comprobación del límite:
lim[n = oo][ | ( (a_{k}·n^{k}+...(k)...+a_{0})/(a_{k}·oo^{k}) )+(-1) | ] = ...
... lim[n = oo][ | ( a_{k}·n^{k}/a_{k}·oo^{k} )+(-1) | ] = 0 [< (1/n) < (1/n_{0}) < s
Teorema:
lim[n = oo][ a_{k}·(1/n)^{k}+...(k)...+a_{0} ] = a_{0}
Demostración:
Sea s > 0 ==>
Se define n_{0} > (1/s)
lim[n = oo][ | ( a_{k}·(1/n)^{k}+...(k)...+a_{0} )+(-1)·a_{0} | = | a_{0}+(-1)·a_{0} | = 0 [< ...
... (1/n) < (1/n_{0}) < s
Teorema:
lim[n = oo][ (an^{2}+bn+c)/(an+b) ] = oo
lim[n = oo][ (an+b)/(an^{2}+bn+c) ] = 0
Definición:
[Ax][Ay][Es][ ( xn^{y+s} <=[R]=> (y+s)·n^{x} ) <==> ( (y+s)·n^{x} <=[R]=> xn^{y+s} ) ]
Teorema:
Sea s = x+(-y) ==>
xn^{x} <=[R]=> xn^{x}
Teorema:
[Eu][ xn^{y+u} <=[R]=> (y+u)·n^{x} ] <==> [Ev][ yn^{x+v} <=[R]=> (x+v)·n^{y} ]
Demostración:
Se define u = 0 & v = 0
Teorema:
Si ( [Eu][ xn^{y+u} <=[R]=> (y+u)·n^{x} ] & [Ev][ yn^{z+v} <=[R]=> (z+v)·n^{y} ] ) ==> ...
... [Ew][ xn^{z+w} <=[R]=> (z+w)·n^{x} ]
Demostración:
Se define u = 0 & v = x+(-z) & w = y+(-z)
( p(x) & ( p(x) <==> q(y) ) ) <==> ( p(x) & q(y) )
( q(y) & q(y) ) <==> q(y)
Teorema:
lim[n = oo][ (3n^{4}+4n^{3}+n^{2}+2n)/(2n^{3}+3n^{2}+1) ] = (3/2)·oo
lim[n = oo][ (2n^{3}+3n^{2}+1)/(3n^{4}+4n^{3}+n^{2}+2n) ] = (2/3)·0
Teorema:
lim[n = oo][ (2n^{3}+3n^{2}+1)/(n^{3}+3n+2) ] = 2
lim[n = oo][ (n^{3}+3n+2)/(2n^{3}+3n^{2}+1) ] = (1/2)
Teorema:
Si a_{n} es de Cauchy ==> a_{n} está acotada
Demostración:
Sea s > 0 ==>
Sea m = n_{0}+1 ==>
|a_{n}| [< |a_{n}+(-1)·a_{m}|+|a_{m}| = |a_{n}+(-1)·a_{n_{0}+1}|+|a_{n_{0}+1}|
Se define M = max{a_{1},...,a_{n_{0}},s+|a_{n_{0}+1}|}
Sea n€N ==>
|a_{n}| [< M
Teorema:
Si a_{n} > 0 ==> lim[n = oo][ a_{n}] >] 0
Si a_{n} < 0 ==> lim[n = oo][ a_{n}] [< 0
Demostración:
Sea s > 0 ==>
a_{n} > 0
a_{n}+(-a) > (-a)
s > a_{n}+(-a) > (-a)
s > (-a)
0 >] (-a)
a >] 0
Ley: [ de caminar sobre las aguas anulando la gravedad de la Tierra ]
[Ef][ < f: A ---> B & x --> f(x) = qg > & m·d_{tt}^{2}[y(t)] = (-1)·qg+f(x) ]
Ley: [ de levantar las aguas anulando la gravedad de la Luna ]
[Ef][ < f: A ---> B & x --> f(x) = (-1)·pg > & m·d_{tt}^{2}[y(t)] = pg+f(x) ]
Ley: [ de volar anulando la gravedad de la Tierra ]
[Ef][ < f: A ---> B & x --> f(x) = qg+F·at > & m·d_{tt}^{2}[y(t)] = (-1)·qg+f(x) ]
Ley: [ de bucear anulando la gravedad de la Luna ]
[Ef][ < f: A ---> B & x --> f(x) = (-1)·pg+(-1)·F·at > & m·d_{tt}^{2}[y(t)] = pg+f(x) ]
Constante de Euler:
Teorema:
lim[n = oo][ ln(n)+(-1)·( 1+...(n)...+(1/n) ) ] = ln(2)
oo·ln(2)+(-1)·oo·ln(2) = ln(2)
Arte: [ de Euler ]
[En][ sum[k = 1]-[n][ ln(1+k)+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+k)+(-1)·(1/k) ] = ln(2)
[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+(1/k))+(-1)·(1/k) ] = ln(2)
Arte:
[En][ sum[k = 1]-[n][ ln(1+k)+(-1)^{k}·(n/k)·(1/2)^{k+(-1)} ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+k)+(-1)^{k}·(oo/k)·(1/2)^{k+(-1)} ] = ln(2)
[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)^{k}·(n/k)·(1/2)^{(1/k)+(-1)} ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+(1/k))+(-1)^{k}·(oo/k)·(1/2)^{(1/k)+(-1)} ] = ln(2)
Arte:
[En][ sum[k = 1]-[n][ ln(1+k)+(-1)^{k}·(n/k)·( 2k+(-1) ) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+k)+(-1)^{k}·(oo/k)·( 2k+(-1) ) ] = ln(2)
[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)^{k}·(1/k)·( 2·(1/k)+(-1) ) ] ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+(1/k))+(-1)^{k}·(oo/k)·( 2·(1/k)+(-1) ) ] ] = ln(2)
Arte:
[En][ sum[k = 1]-[n][ ln(1+k)+(-1)^{k}·(1/k)·[ n // k ] ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+k)+(-1)^{k}·(1/k)·[ oo // k ] ] = ln(2)
[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)^{k}·(1/k)·[ n // (1/k) ] ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ ln(1+(1/k))+(-1)^{k}·(1/k)·[ oo // (1/k) ] ] = ln(2)
Arte:
[En][ lim[s = 1][ sum[k = 1]-[n][ int[ ( 1/(s+1)^{k} ) ]d[s]+(-1)·(1/k)^{s} ) ] = ln(2)+(-1)·(1/n) ]
lim[s = 1][ sum[k = 1]-[oo][ int[ ( 1/(s+1)^{k} ) ]d[s]+(-1)·(1/k)^{s} ) ] ] = ln(2)
[En][ lim[s = 1][ sum[k = 1]-[n][ int[ ( 1/(s+1)^{(1/k)} ) ]d[s]+(-1)·(1/k)^{s} ) ] = ln(2)+(-1)·(1/n) ]
lim[s = 1][ sum[k = 1]-[oo][ int[ ( 1/(s+1)^{(1/k)} ) ]d[s]+(-1)·(1/k)^{s} ) ] ] = ln(2)
Arte:
[En][ sum[k = 1]-[n][ int[x = 1]-[2][ (1/2)·( k·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = 1]-[2][ (1/2)·( k·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)
[En][ sum[k = 1]-[n][ int[x = 1]-[2][ (1/2)·( (1/k)·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = 1]-[2][ (1/2)·( (1/k)·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)
Arte:
[En][ sum[k = 1]-[n][ ...
... int[x = e]-[e^{2}][ ( 1/( k·ln(x) ) )·(1/x)^{k} ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = e]-[e^{2}][ ( 1/( k·ln(x) ) )·(1/x)^{k} ]d[x]+(-1)·(1/k) ] = ln(2)
[En][ sum[k = 1]-[n][ ...
... int[x = e]-[e^{2}][ ( 1/( (1/k)·ln(x) ) )·(1/x)^{(1/k)} ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = e]-[e^{2}][ ( 1/( (1/k)·ln(x) ) )·(1/x)^{(1/k)} ]d[x]+(-1)·(1/k) ] = ln(2)
Arte:
Sea s != (-1) ==>
[En][ sum[k = 1]-[n][ ...
... int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...
... (s+1)·( k·ln(x) )^{s}·( 1/x^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...
... (s+1)·( k·ln(x) )^{s}·( 1/x^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)
[En][ sum[k = 1]-[n][ ...
... int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...
... (s+1)·( (1/k)·ln(x) )^{s}·( 1/x^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...
... (s+1)·( (1/k)·ln(x) )^{s}·( 1/x^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)
Arte:
Sea s != 0 ==>
[En][ sum[k = 1]-[n][ ...
... int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...
... e^{x^{k}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...
... ( k·ln(x) )^{s}+s·( k·ln(x) )^{s+(-1)}·(1/x)^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...
... e^{x^{k}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...
... ( k·ln(x) )^{s}+s·( k·ln(x) )^{s+(-1)}·(1/x)^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)
[En][ sum[k = 1]-[n][ ...
... int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...
... e^{x^{(1/k)}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...
... ( (1/k)·ln(x) )^{s}+s·( (1/k)·ln(x) )^{s+(-1)}·(1/x)^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]
sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...
... e^{x^{(1/k)}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...
... ( (1/k)·ln(x) )^{s}+s·( (1/k)·ln(x) )^{s+(-1)}·(1/x)^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)
Teorema:
Si se explica un teorema entonces se está dentro de los métodos de demostración
porque sinó no sirve para nada explicar, porque no hay amor, fuera de los métodos de demostración.
Demostración:
Se explica un teorema y no se está dentro de los métodos de demostración
aunque no-obstante sirve para algo explicar, porque hay amor, dentro de los métodos de demostración.
Teorema:
lim[s = 0][ int-[m]-[z = se^{ix}+a][ ( f(z)/(z+(-a)) ) ]d[z] ] = (2pi·i)·m·f(a)
lim[s = 0][ int-[m]-[z = se^{ix}+(-a)][ ( f(z)/(z+a) ) ]d[z] ] = (2pi·i)·m·f(-a)
Teorema:
lim[s = 0][ int-[m]-[z = se^{ix}+a][ ( f(z)/( (z+(-a))·(z+a) ) ) ]d[z] ] = (pi·i)·m·(f(a)/a)
lim[s = 0][ int-[m]-[z = se^{ix}+(-a)][ ( f(z)/( (z+(-a))·(z+a) ) ) ]d[z] ] = (pi·i)·m·( f(-a)/(-a) )
Teorema:
lim[s = 0][ int-[m]-[z = ( se^{ix}+a )^{(1/p)}][ ( f(z)/( (z^{p}+(-a))·(z^{q}+a) ) ) ]d[z] ] = ...
... (2pi·i)·m·( f(a)/( a^{(q/p)}+a ) )·(1/p)·a^{(1/p)+(-1)}
lim[s = 0][ int-[m]-[z = ( se^{ix}+(-a) )^{(1/q)}][ ( f(z)/( (z^{p}+(-a))·(z^{q}+a) ) ) ]d[z] ] = ...
... (2pi·i)·m·( f(-a)/( (-a)^{(p/q)}+(-a) ) )·(1/q)·(-a)^{(1/q)+(-1)}
Teorema:
lim[s = 0][ int-[m]-[z = (1/p)·ln( se^{ix}+a )][ ( f(z)/( (e^{pz}+(-a))·(e^{qz}+a) ) ) ]d[z] ] = ...
... (2pi·i)·m·( f(a)/( a^{(q/p)}+a ) )·(1/p)·(1/a)
lim[s = 0][ int-[m]-[z = (1/q)·ln( se^{ix}+(-a) )][ ( f(z)/( (e^{pz}+(-a))·(e^{qz}+a) ) ) ]d[z] ] = ...
... (2pi·i)·m·( f(-a)/( (-a)^{(p/q)}+(-a) ) )·(1/q)·(1/(-a))
Uzkatzen-ten-dut-zû-tek a la gentotzak,
parlatzi-ten-dut-zare-dut en Euskera-Bascotzok parlatzi-koak.
Veurtu-ten-dut-zû-tek a la gentotzak,
escrivitzi-ten-dut-zare-dut en Euskera-Bascotzok parlatzi-koak.
Sepjjakin-ten-dut-zû-tek per les meuotzaks orelli-koaks,
que entendertu-ten-dut-zû-tek el Euskera-Bascotzok parlatzi-koak.
Sepjjakin-ten-dut-zû-tek per els meuotzoks ur-ulli-koaks,
que llezkatzen-ten-dut-zû-tek el Euskera-Bascotzok parlatzi-koak.
Juan:
Hemos sorprendido a esta mujer cometiendo adulterio.
Tú que dices maestro?
Si la mujer está libre de pecado,
que le tiren la primera piedra,
porque tiene condenación,
en no ser pecadora.
Hemos sorprendido a esta hombre cometiendo adulterio.
Tú que dices maestra?
Si el hombre no está libre de pecado,
que no le tiren la primera piedra,
porque no tiene condenación,
en ser pecador.
Carta de Jûanat-hád a los cristianos stronikianos:
Un infiel puede cometer adulterio,
porque no tiene que nacer,
y no está libre de pecado,
porque es un pecador,
y siempre puede haber porno de infieles.
Un fiel no puede cometer adulterio,
porque tiene que nacer,
y está libre de pecado,
porque no es un pecador,
y nunca puede haber porno de fieles.
Sabéis que no nacen los que han cometido adulterio
y entonces también si eran infieles los que han muerto,
es irrelevante,
y alegre en todo-algún caso.
Quizás sabéis que no nacen los que han cometido adulterio
pero si eran fieles los que han muerto,
es relevante,
y triste en todo caso.
No se puede odiar a un fiel,
no cometiendo adulterio,
no follando estando divorciado.
Se puede odiar a un infiel,
cometiendo adulterio,
follando estando divorciado.
Definición: [ de la suma de Riemman ]
[As][ s > 0 ==> [En_{0}][An][ n > n_{0} ==> ...
... | sum[k = 1]-[k = n][ f(k/n)·(1/n) ]+(-1)·int[x = 0]-[1][ f(x) ]d[x] | < s ] ]
Teorema:
int[x = 0]-[1][ e^{x} ]d[x] = ...
... lim[n = oo][ ( e^{(1/n)}+...(n)...+e^{(n/n)} )·( e^{(1/n)}+(-1) ) ] = e+(-1)
Demostración:
lim[n = oo][ e^{(1/n)}+(-1) ] = lim[n = oo][ (1/n)+sum[k = 1]-[oo][ (1/(k+1)!)·(1/n)^{k+1} ] ] = ...
... lim[n = oo][ (1/n) ] = 0
lim[n = oo][ ( e^{(1/n)}+...(n)...+e^{(n/n)} ) ] = ...
... lim[n = oo][ ( (e^{(n/n)+(1/n)}+(-1)·e^{(1/n)})/(e^{(1/n)}+(-1)) ) ]
Teorema:
int[x = 0]-[1][ e^{(-x)} ]d[x] = ...
... lim[n = oo][ ( e^{(-1)·(1/n)}+...(n)...+e^{(-1)·(n/n)} )·( 1+(-1)·e^{(-1)·(1/n)} ) ] = 1+(-1)·(1/e)
Demostración:
lim[n = oo][ 1+(-1)·e^{(-1)·(1/n)} ] = ...
... lim[n = oo][ (1/n)+sum[k = 1]-[oo][ (-1)^{k}·(1/(k+1)!)·(1/n)^{k+1} ] ] = ...
... lim[n = oo][ (1/n) ] = 0
lim[n = oo][ ( e^{(-1)·(1/n)}+...(n)...+e^{(-1)·(n/n)} ) ] = ...
... lim[n = oo][ ( (e^{(-1)·(1/n)}+(-1)·e^{(-1)·( (n/n)+(1/n) )})/(1+(-1)·e^{(-1)·(1/n)}) ) ]
Teorema:
int[x = 0]-[1][ ln(x+1) ]d[x] = ...
... lim[n = oo][ ( ln((1/n)+1)+...(n)...+ln((n/n)+1) )·(1/n) ] = 2·ln(2)+(-1)
Demostración: [ por Stolz ]
f(1) = ( (2n+(-1))/n )
[En][ Id(1) = ( (2n+(-1))/n ) & n = 1 ]
g(0) = ( (1+(-n))/n )
[En][ Id(0) = ( (1+(-n))/n ) & n = 1 ]
lim[n = oo][ ln( ( (n+1)/(n+1) )+1 ) ] = lim[n = oo][ f(1)·ln( ( (n+1)/(n+1) )+1 )+g(0) ] = ...
... lim[n = oo][ ( ( (2n+(-1))/n ) )·ln( ( (n+1)/(n+1) )+1 )+( (1+(-n))/n ) ] = 2·ln(2)+(-1)
Teorema:
int[x = 0]-[1][ cos(x) ]d[x] = ...
... lim[n = oo][ ( cos(1/n)+...(n)...+cos(n/n) )·(1/n) ] = sin(1)
Demostración: [ por Stolz ]
cos( x+(-1)·(pi/2) ) = cos(x)·cos( (-1)·(pi/2) )+(-1)·sin(x)·sin( (-1)·(pi/2) ) = sin(x)
f(0) = x·(pi/2)
g(x) = (-1)
lim[n = oo][ cos(k/(n+1))+(-1)·cos(k/n) ] = cos(0)+(-1)·cos(0) = 1+(-1) = 0
lim[n = oo][ cos((n+(-k))/(n+1))+(-1)·cos((n+(-k))/n) ] = cos(1)+(-1)·cos(1) = 0
lim[n = oo][ cos((n+1)/(n+1)) ] = lim[n = oo][ cos( (n+1)/(n+1)+f(0) ) ] = ...
... lim[n = oo][ cos( (n+1)/(n+1)+x·(pi/2) ) ] = lim[n = oo][ cos( (n+1)/(n+1)+g(x)·(pi/2) ) ] = ...
... lim[n = oo][ cos( (n+1)/(n+1)+(-1)·(pi/2) ) ] = lim[n = oo][ sin( (n+1)/(n+1) ) ] = sin(1)
Examen de sumas de Riemman:
Teorema:
int[x = 0]-[1][ sin(x) ]d[x] = ...
... lim[n = oo][ ( sin(1/n)+...(n)...+sin(n/n) )·(1/n) ] = 1+(-1)·cos(1)
Demostración: [ por Stolz ]
sin( x+(-1)·(pi/2) ) = sin(x)·cos( (-1)·(pi/2) )+cos(x)·sin( (-1)·(pi/2) ) = (-1)·cos(x)
f(0) = x·(pi/2)
g(x) = (-1)
h(0) = ( (n+(-1))/n )
Teorema:
sinh(ix+iy) = ( sin(x+y)/i ) = (1/i)·( sin(x)·cos(y)+cos(x)·sin(y) ) = sinh(ix)·cosh(iy)+cosh(ix)·sinh(iy)
cosh(ix+iy) = cos(x+y) = ( cos(x)·cos(y)+(-1)·sin(x)·sin(y) ) = cosh(ix)·cosh(iy)+sinh(ix)·sinh(iy)
Teorema:
int[x = 0]-[1][ cosh(x) ]d[x] = ...
... lim[n = oo][ ( cosh(1/n)+...(n)...+cosh(n/n) )·(1/n) ] = sinh(1)
Demostración: [ por Stolz ]
cosh( x+i·(pi/2) ) = cosh(x)·cosh( i·(pi/2) )+sinh(x)·sinh( i·(pi/2) ) = i·sinh(x)
f(0) = x·(pi/2)
g(x) = i
h(i) = (1+(-i))·( (n+(-1))/n )+i
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